Angular Frequency of SHM for Rolling Disk with Springs

To find the angular frequency $\omega$ of the simple harmonic motion (SHM) performed by the centre of mass of the disk, we can use the energy method or the force-torque method.

Method 1: Energy Conservation Method

Let $x$ be the displacement of the centre of mass of the disk from its equilibrium position, and $v = \frac{dx}{dt}$ be its linear velocity.

Since the disk rolls without slipping on the horizontal surface, its angular velocity $\omega_{\text{rot}}$ about its centre of mass is related to $v$ by:
$\omega_{\text{rot}} = \frac{v}{R}$

The total mechanical energy $E$ of the system consists of the kinetic energy of rolling and the potential energy stored in the two springs:
$E = E_{\text{kinetic}} + U_{\text{spring}}$

1. Kinetic Energy ($E_{\text{kinetic}}$):
   $E_{\text{kinetic}} = \frac{1}{2} M v^2 + \frac{1}{2} I_{\text{CM}} \omega_{\text{rot}}^2$

   For a uniform thin cylindrical disk, the moment of inertia about its central axis is $I_{\text{CM}} = \frac{1}{2} M R^2$. Substituting $I_{\text{CM}}$ and $\omega_{\text{rot}}$:
   $E_{\text{kinetic}} = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2$

2. Potential Energy ($U_{\text{spring}}$):
   Since there are two identical springs each of spring constant $k$:
   $U_{\text{spring}} = 2 \times \left( \frac{1}{2} k x^2 \right) = k x^2$

Thus, the total energy of the system is:
$E = \frac{3}{4} M v^2 + k x^2$

Since the system is conservative, the total mechanical energy $E$ remains constant with time ($\frac{dE}{dt} = 0$):
$\frac{dE}{dt} = \frac{d}{dt} \left( \frac{3}{4} M v^2 + k x^2 \right) = 0$

$\frac{3}{4} M \cdot \left(2v \frac{dv}{dt}\right) + k \cdot \left(2x \frac{dx}{dt}\right) = 0$

Since $v = \frac{dx}{dt}$ and acceleration $a = \frac{dv}{dt}$:
$\frac{3}{2} M v a + 2 k x v = 0$

Dividing throughout by $v$ ($v \neq 0$):
$\frac{3}{2} M a + 2 k x = 0 \implies a = -\left(\frac{4k}{3M}\right) x$

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Method 2: Force and Torque Equations

For a displacement $x$ towards the right:

1. The combined force exerted by both springs is $F_{\text{spring}} = -2kx$ (acting towards the left).
2. Let $f$ be the force of static friction acting on the disk at the point of contact with the floor to prevent slipping, directed towards the left.

Writing Newton's second law for translational motion:
$-2kx - f = M a \quad \text{--- (1)}$

Writing the torque equation about the centre of mass for rotation:
$\tau_{\text{CM}} = I_{\text{CM}} \alpha$
$f R = \left( \frac{1}{2} M R^2 \right) \alpha$

For pure rolling, $a = R \alpha$, so:
$f R = \frac{1}{2} M R^2 \left( \frac{a}{R} \right) \implies f = \frac{1}{2} M a \quad \text{--- (2)}$

Substituting equation (2) into equation (1):
$-2kx - \frac{1}{2} M a = M a$
$-2kx = \frac{3}{2} M a \implies a = -\left(\frac{4k}{3M}\right) x$

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Conclusion

Comparing the acceleration expression $a = -\left(\frac{4k}{3M}\right) x$ with the standard differential equation for Simple Harmonic Motion ($a = -\omega^2 x$), we get:

$\omega^2 = \frac{4k}{3M} \implies \omega = \sqrt{\frac{4k}{3M}}$

Therefore, the correct option is D.